The Principle of Σ 45... ( needs work and spell che

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The Principle of Σ 45... ( needs work and spell che

Postby StrmCkr » Tue Sep 12, 2006 11:17 pm

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Postby ronk » Wed Sep 13, 2006 12:57 am

needs work and spell che...

Your puzzle needs work too. It has 5 solutions.
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Postby StrmCkr » Wed Sep 13, 2006 3:20 am

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Postby daj95376 » Wed Sep 13, 2006 4:00 am

I guess Fiendish puzzles aren't what they use to be. You went to all this effort to solve a puzzle with one Naked Pair and a ton of Singles!
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Postby StrmCkr » Wed Sep 13, 2006 4:11 am

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Postby udosuk » Wed Sep 13, 2006 8:43 am

StrmCkr wrote:Step Eight:
Eliminate possible solutions by breaking a number in a given solution down into its smaller components(if possible) and see what number remains.

Limitation Check #1: A number cannot repeat ever! Breaking it into smaller parts to prove a specific solution is not the correct one to use.

(5+2)
((3+2) + 2);

This cannot be used because it violates the limitation of the puzzle in two ways. The number “3” is known and cannot repeat, the number “2” is unknown and cannot repeat twice.

This also proves that the number 5 is not the solution because it cannot be broken into simple parts and itself is not found used in the puzzle so some other combination of missing numbers equal the sum of 5. This proves that the solution of 5+2 ≠ 7. and there for cannot be used.

Can you elaborate more why A cannot be 5 and must be 2? Of course, a naked triple in box 4 will force r4c3=5, but I just can't follow your ways of counting the 45-sum to conclude that A (r4c7) must not be 5...:(

Just so you know, this concept of 45-rule is used excessively in killer sudoku/Sum Number Place/Samunamupure... You might like to try them if you're bored of the vanilla sudoku...
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Postby StrmCkr » Wed Sep 13, 2006 6:10 pm

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Postby udosuk » Wed Sep 13, 2006 7:24 pm

StrmCkr wrote:we know that 3 is a used number, this implies that the # 5 can only be determined / is dependent on the # 2. Then i take this knowledge back into the inital equation.

(3+2)+2 and discover that the "2" becomes a repeating number in this equation creating a false solution. (thanks to the rules of the puzzle #'s cannot repeat).

there for

(5+2) ≠ 7

and this leaves

7| (4+2+1) as the only valid solution remaining and the only unknown # in this solution is 2
A = 2

This is the bit bothering me... I guess you're manipulating the candidate list of cell A (r4c7)... Obviously the sum of candidates in A must be 7 (Simple Sudoku determined that the candidate list must be {2,5})...

So, what you did, first you determined that the list must be either {2,5} or {1,2,4}, but for some reason you thought that {2,5} must be further broken up (WHY?:?: ) And must be either {2,{1,4}} or {2,{2,3}}, which were both impossible... But what about the original {2,5}!:!:

Back to the start... A=(45)-(9+8+7+6+4+3+1)... Here you're eliminating 7 different candidates out of a total of 9 possibilities... That shows you must have 2 possible condidates left, which sums to 7: {1,6}, {2,5} or {3,4}. But {1,6} and {3,4} all appeared on the cells which A could see directly, therefore the only possibility is {2,5}.

BUT you cannot eliminate 5 from cell A without looking at other cells!!!

StrmCkr wrote:(5+2) ≠ 7

And this is the most bizarre statement I've seen all year...:(
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Postby StrmCkr » Wed Sep 13, 2006 7:52 pm

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Postby udosuk » Wed Sep 13, 2006 9:01 pm

StrmCkr wrote:does that make n e sence?
Sorry I'm afraid not... At least not to me...:(

I suspect what you're doing here is similar to the "Uniqueness principles", but my gut feeling tells me there's some fallacy within your reasoning...

StrmCkr wrote:if "a" saw a 2 then all solutions of the number 5 would be seen and then number 5 would become the only number that could equal that score . because the 2 is not seen "5" must be a a sum of other numbers.

This is like saying "if George W. Bush was to be USA president for another 4 years then USA would invade Iran, but since he's gonna retire in 2 years, USA must not invade Iran, ever.

Of course we all hope this turns out to be true, but logic says there's something wrong with this reasoning, for there could be another president, be it Al Gore, Hillary Clinton or Arnold Schwarzenegger, who could decide to invade Iran for various reasons...

Similarly, of course if cell A sees a "2" then the only candidate left for it must be "5"... But now that we know A doesn't see a "2", it could still be "5", instead of a combination of {1,4} or {2,3}... You have to find other reasons to support your claim that cell A cannot be "5"... Don't tell me "it's hard to visualise" or something alike... Just tell me explicitly, what could go wrong if cell A doesn't see a "2" and cell "A" is in fact "5"...
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Postby StrmCkr » Wed Sep 13, 2006 10:49 pm

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Postby StrmCkr » Fri Sep 15, 2006 7:53 am

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Postby Mike Barker » Fri Oct 13, 2006 1:34 pm

I'm trying to get my arms around this technique. Maybe this will help. If I take the last example and exchange 9 with 2 and vice versa I end up with:
Code: Select all
. 1 .|. . .|. 5 .
7 . 8|. . .|. . .
. . 9|6 . 8|3 . .
-----+-----+-----
2 . .|3 . 5|. . 7
. . .|. . .|. . .
8 . .|4 . 9|. . 6
-----+-----+-----
. . 2|7 . 1|9 . .
. . 3|. . .|4 . .
. 8 .|. . .|. 2 .

The question is, "how does the logic of the above post change with this new puzzle?" My problem is that I don't see any differences, which would lead to the conclusion that A4=2 which is incorrect.
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Postby StrmCkr » Wed Nov 22, 2006 8:41 am

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Postby Mike Barker » Wed Nov 22, 2006 2:09 pm

Thanks for the reply. I enjoy seeing new techniques. I haven't had time to go through everything you've posted, but two questions. I can see one ALC which I interpret to be an Almost Locked Set: a4,b4,j4=1259, but what is the second ALS involving e4=128? Also the pencil marks for your original grid are:
A4 = 2,9
b4= 1,2,5,9
e4 = 1,8,9
h4 = 2,5,8,9
j4 = 2,5
(Notice the difference in e4) so are your conclusions regarding the differences between the two puzzles still valid?
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